Integration by parts

$ \displaystyle\int u dv = uv - \int v du$

  (a) $\displaystyle \int \sqrt{x}\ln x dx = \frac{2}{3} x^\frac{3}{2} \ln x - \frac{...
...e{0.8in}\fbox{(B) $u = \ln x$,\: $v = \frac{2}{3} x^\frac{3}{2}$}\hspace{5.0in}$    
  $\displaystyle \hspace{0.4in} = \frac{2}{3} x^\frac{3}{2} \ln x - \frac{4}{9} x^\frac{3}{2} + C$    

  (b) $\displaystyle \int e^x\sin x dx = e^x \sin x - \int e^x\cos x dx \hspace{0.8in}\fbox{(B) $u = \sin x$,\: $v = e^x$}$   (Maple cannot simplify.)$\displaystyle \hspace{5.0in}$    
  $\displaystyle \hspace{0.4in} = e^x \sin x - \left[e^x\cos x + \int e^x\sin x dx\right] \hspace{0.8in}\fbox{(B) $u = \cos x$,\: $v = e^x$}\hspace{5.0in}$    
  $\displaystyle \hspace{0.4in} = e^x \sin x - e^x\cos x - \int e^x\sin x dx,$   $\displaystyle \mbox{ which implies $\displaystyle\int e^x\sin x dx = \frac{e^2}{2}(\sin x - \cos x)$. }$    

  (c) $\displaystyle \int x^3 e^{-x^2} dx = \frac{1}{2}\int t e^{-t} dt \hspace{0.8in}\fbox{(A.1) $t = x^2$,\: $\frac{dt}{dx} = 2x$}\hspace{5.0in}$    
  $\displaystyle \hspace{0.4in} = \frac{1}{2}\left[-t e^{-t} + \int e^{-t} dt\right] \hspace{0.8in}\fbox{(B) $u = t$,\: $v = -e^{-t}$}\hspace{5.0in}$    
  $\displaystyle \hspace{0.4in} = \frac{1}{2}\left[-t e^{-t} + \int e^{-t} dt\rig...
...-t} - e^{-t}\right] + C = - \frac{1}{2} x^2 e^{-x^2} - \frac{1}{2} e^{-x^2} + C$    

  (d) $\displaystyle \int \frac{x}{2 + e^x + e^{-x}} dx = 4 \int \frac{t}{(e^t + e^{-...
...2} dt \hspace{0.8in}\fbox{(A.2) $x = 2t$,\: $\frac{dx}{dt} = 2$}\hspace{5.0in}$    
  $\displaystyle \hspace{0.4in} = \int \frac{t}{\cosh^2 t} dt = t \tanh t - \int \tanh t dt \hspace{0.8in}\fbox{(B) $u = t$,\: $v = \tanh t$}\hspace{5.0in}$    
  $\displaystyle \hspace{0.4in} = t \tanh t - \int \frac{\sinh t}{\cosh t} dt = t...
...ace{0.8in}\fbox{(A.1) $s = \cosh t$,\: $\frac{ds}{dt} = \sinh t$}\hspace{5.0in}$    
  $\displaystyle \hspace{0.4in} = t \tanh t - \ln \vert s\vert + C = (x/2) \tanh (x/2) - \ln \vert\cosh(x/2)\vert + C$    

  (e) $\displaystyle \int x^3\sin x^2 dx = \frac{1}{2}\int t\sin t dt \hspace{0.8in}\fbox{(A.1) $t = x^2$,\: $\frac{dt}{dx} = 2x$}\hspace{5.0in}$    
  $\displaystyle \hspace{0.4in} = \frac{1}{2}\left[-t\cos t + \int\cos t dt\right] \hspace{0.8in}\fbox{(B) $u = t$,\: $v = -\cos t$}\hspace{5.0in}$    
  $\displaystyle \hspace{0.4in} = -\frac{t}{2}\cos t + \frac{1}{2}\sin t + C = -\frac{x^2}{2}\cos x^2 + \frac{1}{2}\sin x^2 + C$    

  (f) $\displaystyle \int \sin x \ln\vert\cos x\vert dx = \int -\ln\vert t\vert dt \hspace{0.8in}\fbox{(A.1) $t = \cos x$,\: $\frac{dt}{dx} = -\sin x$}\hspace{5.0in}$    
  $\displaystyle \hspace{0.4in} = -t \ln\vert t\vert + \int dt \hspace{0.8in}\fbox{(B) $u = \ln\vert t\vert$,\: $v = -t$}\hspace{5.0in}$    
  $\displaystyle \hspace{0.4in} = -t \ln\vert t\vert + t + C = -\cos x \ln\vert\cos x\vert + \cos x + C$    

  (g) $\displaystyle \int \frac{x^2}{(x^2 + 8)^\frac{3}{2}} dx = -x(x^2 + 8)^\frac{1}...
... \hspace{0.8in}\fbox{(B) $u = x$,\: $v = -(x^2 + 8)^\frac{1}{2}$}\hspace{5.0in}$    
  $\displaystyle \hspace{0.4in} = -\frac{x}{\sqrt{x^2 + 8}} + \int \frac{1}{\sqrt{...
...{0.8in}\fbox{(A.2) $x = \sqrt{8}t$,\: $\frac{dx}{dt} = \sqrt{8}$}\hspace{5.0in}$    
  $\displaystyle \hspace{0.4in} = -\frac{x}{\sqrt{x^2 + 8}} + \sinh^{-1} t + C = -\frac{x}{\sqrt{x^2 + 8}} + \sinh^{-1}\left(\frac{x}{\sqrt{8}}\right) + C$    



Department of Mathematics
Last modified: 2005-10-05